y = x^99 +99^x.
To determine whether y(x) is increasing or decreasing.
Even by common sense we a function x , x^3 or x^99 is an incresing function. The secod term 99^x is an exponential function of a positive 99 is also increasing . So the sum is an increasing.
Now we see by calculus:
We know that for an increasing function the derivative y'(x) should be positive .
Therefore differentiating the given function we get:
y'(x) = (x^99+99^x)'
y'(x) = 99x^98 + (99^x) log 99. Both terms are positive for all x . Therefore y(x) is increasing.
In order to prove that f(x) is an increasing function, we'll do the first derivative test.
If the first derivative of the function is positive, then the function is increasing.
If the first derivative of the function is negative, then the function is decreasing.
Let's calculate f'(x):
f'(x) = (x^99 + 99^x)'
f'(x) = (x^99)' + (99^x)'
f'(x) = 99x^98 + 99^x*ln99
Since x^98 is positive for any real value of x, then 99x^98 is also positive.
We'll calculate ln99 = 4.595 approx. (1)
99^x >0 (2)
From (1) and (2) => 99^x*ln99>0
f'(x)=99x^98 + 99^x*ln99>0
Since each term of the expression of f'(x) is positive, the sum of positive terms is also a positive expression. The expression of f'(x) it's obviously>0, so f(x) is an increasing function.